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Is the angular velocity of a rigid body about any point the same as that about the axis of rotation. Also, can we even define angular terms (Angular Velocity, Angular Acceleration, etc) about any axis, other than the axis of rotation?

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Given some central point $\vec c_1$, the velocity of this point $\vec v_{c_1}$ with respect to some frame of reference, and an angular velocity $\vec\omega$, the velocity of some other point $\vec x$ on a rigid body is $\vec v_x = \vec v_{c_1} + \vec\omega \times (\vec x - \vec c_1)$.

What if some other point $\vec c_2$ is chosen as the central point? The expression for the velocity of the point $\vec x$ becomes $\vec v_x = \vec v_{c_2} + \vec\omega \times (\vec x - \vec c_2)$, where $\vec v_{c_2} = \vec v_{c_1} + \vec\omega \times (\vec c_2 - \vec c_1)$. The angular velocity does not change. It is the same regardless of which point one chooses as the central point. Angular velocity is a free vector.


Update, because the above apparently isn't satisfactory to some

A rigid body is an object for which a frame of reference exists such that the location of every point on the rigid body is constant from the perspective of this frame. In other words, $(\frac {d\boldsymbol x}{dt})_F = 0$ for every point $x$ in the rigid body, with the derivative taken from the perspective of the fixed frame $F$.

Suppose at some time $t$ you know the location $\boldsymbol X(\boldsymbol c,t)$ of some fixed point $\boldsymbol c$ on the rigid body in some other frame of reference $I$. The location of the point $\boldsymbol x$ in this other frame is related to the location of the point $\boldsymbol c$ via $$\boldsymbol X(\boldsymbol x,t) = \boldsymbol X(\boldsymbol c,t) + \boldsymbol{\mathrm R}_{F\to I}(t)(\boldsymbol x - \boldsymbol c) \tag{1}$$ where $\boldsymbol{\mathrm R}_{F\to I}(t)$ is the rotation matrix that transforms coordinates from frame $F$ to frame $I$.

Suppose you know some other point $\boldsymbol c'$, also fixed with respect to the rigid body. The location of the point $\boldsymbol c'$ in the non-fixed frame is $$\boldsymbol X(\boldsymbol c',t) = \boldsymbol X(\boldsymbol c,t) + \boldsymbol{\mathrm R}_{F\to I}(t)(\boldsymbol c' - \boldsymbol c)\tag2$$ The location of the point $\boldsymbol x$ in the non-fixed frame can be re-expressed to be in terms of $\boldsymbol c'$: $$\boldsymbol X(\boldsymbol x,t) = \boldsymbol X(\boldsymbol c',t) + \boldsymbol{\mathrm R}_{F\to I}(t)(\boldsymbol x - \boldsymbol c') \tag{3}$$ Differentiating equations (1) and (3) with respect to time yields $$\dot {\boldsymbol X}(\boldsymbol x,t) = \dot {\boldsymbol X}(\boldsymbol c,t) + \dot{\boldsymbol{\mathrm R}}_{F\to I}(t)(\boldsymbol x - \boldsymbol c) = \dot{\boldsymbol X}(\boldsymbol c',t) + \dot {\boldsymbol{\mathrm R}}_{F\to I}(t)(\boldsymbol x - \boldsymbol c') \tag4$$ The time derivative of a transformation matrix $\boldsymbol{\mathrm R}(t)$ from one Cartesian frame to another can be written as a product of that matrix and a skew symmetric matrix: $$\dot{\boldsymbol{\mathrm R}}(t) = \boldsymbol{\mathrm R}(t) \boldsymbol{\mathrm S}(t)$$ This is valid for a Euclidean space of any dimensionality. In three dimensional space (and three dimensional space only), such a skew symmetric matrix can be mapped to and from a three dimensional pseudo vector: $$\dot{\boldsymbol{\mathrm R}}(t) = \boldsymbol{\mathrm R}(t) \operatorname{Sk}(\boldsymbol\omega(t))$$ Rewriting equation (4) in terms of this pseudo vector, $$\dot {\boldsymbol X}(\boldsymbol x,t) = \dot {\boldsymbol X}(\boldsymbol c,t) + \boldsymbol{\mathrm R}_{F\to I}(t)\left(\boldsymbol\omega\times(\boldsymbol x - \boldsymbol c)\right) = \dot{\boldsymbol X}(\boldsymbol c',t) + \boldsymbol{\mathrm R}_{F\to I}(t)\left(\boldsymbol\omega\times(\boldsymbol x - \boldsymbol c')\right)\tag5$$ Without loss of generality, one can choose a non-fixed frame that is instantaneously co-aligned with the fixed frame at time $t$. With this, equation (5) reduces to $$\dot {\boldsymbol X}(\boldsymbol x,t) = \dot {\boldsymbol X}(\boldsymbol c,t) + \boldsymbol\omega\times(\boldsymbol x - \boldsymbol c) = \dot{\boldsymbol X}(\boldsymbol c',t) + \boldsymbol\omega\times(\boldsymbol x - \boldsymbol c')\tag6$$ This is of course identical to what I originally wrote.

That angular velocity expressed as a skew symmetric matrix is the same regardless of which point one chooses as the origin applies in all Euclidean spaces in which time is the independent parameter of motion. Angular velocity is related to the time derivative of the transformation matrix, and changing origins doesn't change the transformation matrix in an affine transformation (e.g., equations (1) to (3) are affine transformations).

That angular velocity expressed as a pseudo vector is the same regardless of which point one chooses as the origin applies only in three dimensional Euclidean spaces in which time is the independent parameter of motion. That specialization is rather important because we apparently live in a universe that locally appears to be a three dimensional Euclidean space with time as the independent parameter of motion. In other words, we live in a universe where Newtonian mechanics is locally valid. This is the context in which this question was asked and in which I wrote this answer.

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  • $\begingroup$ But isn't $\omega$ defined with respect to some axis? $\endgroup$ – Aaron Stevens Jul 22 '18 at 20:44
  • $\begingroup$ $\omega=\frac{d\theta}{dt}$. How is this independent of the reference axis? $\endgroup$ – Aaron Stevens Jul 22 '18 at 21:08
  • $\begingroup$ @AaronStevens - Angular velocity is not $\frac{d\theta}{dt}$, at least not in our three dimensional world. Angular displacement is not a vector. Angular velocity is a vector (or, more technically, it's a pseudo vector), related closely to the time derivative of the transformation matrix from an inertial frame of reference to a frame of reference fixed with respect to the rigid body in question. $\endgroup$ – David Hammen Jul 22 '18 at 21:25
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    $\begingroup$ @DavidHammen This answer does not attempt to define "angular velocity of a rigid body about any point", which is, I believe, what the question is about. Instead, it just shows how, assuming some angular velocity of a rotating rigid body, the linear velocity of a point on that body could be expressed relative to a random point in space. $\endgroup$ – V.F. Jul 24 '18 at 11:58
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    $\begingroup$ @AaronStevens Yes, in my answer (to avoid the term "center") I use the term "reference point" (p2 on diagram) to refer to an alternative point in space relative to which angular velocity could be defined and measured. If I stay on the ground and watch a plane in the sky, I can measure angular velocity of the plane, where angle between different positions of the plane is measured from my position, which serves as the vertex of an angle. The plane could follow straight line, circle or move randomly. Each point of the plane will have slightly different angular velocity. $\endgroup$ – V.F. Jul 25 '18 at 17:27
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Is the angular velocity of a rigid body about any point the same as that about the axis of rotation?

The angular velocity would not be the same for other reference points.

This is because the axis of rotation is the only locus in space to which any point of a rigid body will maintain the same distance (radius of rotation) throughout the rotation and, therefore, the same ratio between the linear and the angular velocities.

Since the radius of rotation for any other reference point will be changing, the ratio between the linear and angular velocities will be changing as well, so, given the same linear velocities, the angular velocities would have to be different.

Also, can we even define angular terms (Angular Velocity, Angular Acceleration, etc) about any axis, other than the axis of rotation?

According to Wikipedia, "the angular velocity of a particle is the rate at which it rotates around a chosen center point".

If we don't impose any other constraints, there is no obvious reason why angular velocity cannot be measured relative to any fixed point in space. If that point was not fixed, defining and measuring the angle between subsequent radii would become problematic. By that logic, an alternative reference point should not be on the rotating body.

With such definition, instantaneous angular velocity can be measured for any trajectory - not just circular.

Of course, if we interpret the word "center" in the definition as a point that has to remain at the same distance from any given rotating point, we won't be able to define the angular velocity about any other point in space.

Adding a diagram for clarity.

enter image description here

Here $\omega$ is the original, constant, angular velocity of any point of a rotating rigid body defined relative to the center of rotation $p_1$, while $\omega'$ is a variable angular velocity defined for the same point of the rotating rigid body relative to a random fixed point in space, $p_2$. Obviously, $\omega'\ne \omega$.

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  • $\begingroup$ Angular velocity is a free vector. It is the same for all points on a rigid body. $\endgroup$ – David Hammen Jul 22 '18 at 19:52
  • $\begingroup$ @DavidHammen Does it also mean that it should be the same relative to any point in space? $\endgroup$ – V.F. Jul 22 '18 at 19:57
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Jul 24 '18 at 19:46

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